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Question Number 66489 by rajesh4661kumar@gmail.com last updated on 16/Aug/19

Answered by $@ty@m123 last updated on 16/Aug/19

$$x=4$$$$y=9$$

Commented by rajesh4661kumar@gmail.com last updated on 16/Aug/19

$$how$$

Answered by MJS last updated on 16/Aug/19

$$\mathrm{first},\mathrm{try}\mathrm{a}\mathrm{few}\mathrm{values}\mathrm{for}x\mathrm{and}y,\mathrm{both}\mathrm{are}$$$$\mathrm{square}\mathrm{numbers}$$$$\left(1\right)y=11-\sqrt{x}$$$$x=1^2\Rightarrow y=10$$$$x=2^2\Rightarrow y=3^2\left(2\right)2^2+3=7\mathrm{solution}$$$$\mathrm{because}\mathrm{of}\left(2\right)x\mathrm{must}\mathrm{be}\mathrm{smaller}\mathrm{than}7,\mathrm{so}$$$$\mathrm{no}\mathrm{other}\mathrm{integer}\mathrm{solution}\mathrm{is}\mathrm{possible}$$$$$$$$\mathrm{second},\mathrm{without}\mathrm{squaring}(\mathrm{because}\mathrm{this}$$$$\mathrm{would}\mathrm{lead}\mathrm{to}\mathrm{false}\mathrm{solutions})\mathrm{draw}\mathrm{the}$$$$\mathrm{functions}$$$$y=11-\sqrt{x}$$$$x=7-\sqrt{y}$$$$\mathrm{these}\mathrm{are}2\mathrm{half}\mathrm{parabolas}\mathrm{which}\mathrm{only}\mathrm{have}$$$$\mathrm{one}\mathrm{real}\mathrm{intersection}$$$$$$$$\mathrm{third},\mathrm{are}\mathrm{there}\mathrm{complex}\mathrm{solutions}?$$$$\mathrm{no}\mathrm{because}\mathrm{the}\mathrm{full}\mathrm{parabolas}\mathrm{have}4\mathrm{real}$$$$\mathrm{intersections}{\mathrm{there}}^{\prime }\mathrm{s}‘‘\mathrm{no}{\mathrm{place}}^{″}\mathrm{for}\mathrm{other}$$$$\mathrm{solutions}$$

Answered by behi83417@gmail.com last updated on 16/Aug/19

$$(\sqrt{\mathrm{x}}-2)+(\sqrt{\mathrm{y}}-3)(\sqrt{\mathrm{y}}+3)=0$$$$(\sqrt{\mathrm{x}}-2)(\sqrt{\mathrm{x}}+2)+(\sqrt{\mathrm{y}}-3)=0$$$$\Rightarrow -(\sqrt{\mathrm{y}}-3)(\sqrt{\mathrm{y}}+3)(\sqrt{\mathrm{x}}+2)+(\sqrt{\mathrm{y}}-3)=0$$$$\Rightarrow (\sqrt{\mathrm{y}}-3)[-(\sqrt{\mathrm{y}}+3)(\sqrt{\mathrm{x}}+2)+1]=0$$$$\Rightarrow \sqrt{\mathrm{y}}-3=0\Rightarrow \mathrm{y}=9$$$$\sqrt{\mathrm{x}}=11-9=2\Rightarrow \mathrm{x}=4.◼$$

Commented by Prithwish sen last updated on 16/Aug/19

$$\mathrm{but}\mathrm{sir}$$$$\mathrm{x}+\frac{1}{\sqrt{\mathrm{x}}+2}-3=7\Rightarrow \mathrm{x}+\frac{1}{\sqrt{\mathrm{x}}+2}=10$$

Commented by behi83417@gmail.com last updated on 21/Aug/19

$$\sqrt{\mathrm{y}}+3=\frac{1}{\sqrt{\mathrm{x}}+2}\Rightarrow \mathrm{x}+\frac{1}{\sqrt{\mathrm{x}}+2}+3=7$$$$\Rightarrow \mathrm{x}\sqrt{\mathrm{x}}+2\mathrm{x}=4\sqrt{\mathrm{x}}+8\Rightarrow$$$$(\mathrm{x}-4)\sqrt{\mathrm{x}}+2(\mathrm{x}-4)=0\Rightarrow (\mathrm{x}-4)(\sqrt{\mathrm{x}}+2)=0$$$$\stackrel{\sqrt{\mathrm{x}}+2\ne 0}{\Rightarrow }\mathrm{x}-4=0\Rightarrow \mathrm{x}=4.◼$$

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